# when is it true that $x^\prime \Pi \mathbf{1}+ \mathbf{1}^\prime \Pi^\prime x = 2x^\prime \Pi \mathbf{1}$

let $x^\prime$ be a $1 \times n$ vector , $\Pi$ be a $n\times n$ matrix, and $\mathbf{1}^\prime\$ be a $1 \times n$ vector of 1's. When is it true that $x^\prime \Pi \mathbf{1}+ \mathbf{1}^\prime \Pi^\prime x = 2x^\prime \Pi \mathbf{1}$. For example,if $\Pi$ is a $2 \times 2$ symmetric matrix then:
$[x_{1} , x_{2}] \begin{bmatrix} \pi_{1} & \pi_{2} \\ \pi_{2} & \pi_{3} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + [1 , 1] \begin{bmatrix} \pi_{1} & \pi_{2} \\ \pi_{2} & \pi_{3} \end{bmatrix} [x_{1} , x_{2}]^\prime = 2 x_1 (\pi_1 + \pi_2) + x_2 (\pi_2 + \pi_3) = 2x^\prime \Pi \mathbf{1}$

But in general ? Thanks

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## 1 Answer

Why put the same term on $x'\Pi\mathbf1$ on both sides without cancelling? You are just asking whether $x'\Pi\mathbf1=\mathbf1'\Pi'x$, and since both sides are transposes of each other, and $1\times1$ matrices, which are always symmetric, this is true. Even $x'A'y=y'Ax$ holds true whenever $x,y$ are one-column matrices, for the same reason.

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of course. Thanks. –  Luap Nalehw Oct 20 '12 at 17:32